Aluminium, Density, alloys, chemistry, materials!?

Question

The density of pure aluminium calculated from crystallographic data is expected to be 2.6995g/cm³. (can consider density only)

Design an aluminium alloy that has a density of 2.6450g/cm³

and antoher aluminium alloy that has density of 2.7450g/cm³

I'm sort of lost with this question, if someone can help me with the first one and maybe put me on the right track with some formulas or links or what ever. Am i going to have to use trial and error or is there an exact way to do this. Any idea;s would be great!

I'm not looking to actually make something, its more focused on the maths, formula and structure involved, ffc cubic, the mixing of elements, valancey's. Well thats what I think I could be wrong

Answer 1

Most gem stones are aluminum crystals with a little bit of other elements that give the color.
Look up ruby Sapphire and emerald, They may be just what you are looking for and they are the kinds of answers teachers thrive on. You can make ruby with household aluminum foil and acid but learn more about acids before you try to get rich they are very unforgiving to skin.

Answer 2

I think they want you to add another metal to bring the density of the mixture to the specified value. For the lighter than aluminum alloy add magnesium, for the heavier one you could use zinc (these metals are commonly alloyed with aluminum). I'm not certian how they expect you to calculate the quantities unless it is just by weighted percentage of densities (this assumes introduction of other metal induces no crystal structure changes)

density of Mg x fraction of Mg + density of Al x fraction of Al = 2.6450g/cm3

Let x = fraction of magnesuim in the alloy; (thus 1-x is fraction of aluminum)

1.783g/cm3x + 2.6995g/cm3(1-x) = 2.6450g/cm3

1.783x + 2.6995 -2.6995x = 2.6450

-0.9165x + 2.6995 = 2.6450

-0.9165x = - 0.0545

-x = -0.0595

x = 0.0595

So 94.05% aluminum + 5.95% magnesuim This seems to be reasonable, as we only are looking for a slight reduction in density for the new alloy.

(I will leave the heavier alloy for you to do; the math is set up the same)